Integral de $$$\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}$$$

Halla $$$\int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx$$$.

Extrae un seno y expresa todo lo demás en función del coseno, usando la fórmula $$$\sin^2\left(\alpha \right)=-\cos^2\left(\alpha \right)+1$$$ con $$$\alpha=x$$$:

$${\color{red}{\int{\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} d x}}}$$

Sea $$$u=\cos{\left(x \right)}$$$.

Entonces $$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (los pasos pueden verse »), y obtenemos que $$$\sin{\left(x \right)} dx = - du$$$.

La integral puede reescribirse como

$${\color{red}{\int{\left(1 - \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(- u^{2} \left(1 - u^{2}\right)\right)d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = u^{2} \left(1 - u^{2}\right)$$$:

$${\color{red}{\int{\left(- u^{2} \left(1 - u^{2}\right)\right)d u}}} = {\color{red}{\left(- \int{u^{2} \left(1 - u^{2}\right) d u}\right)}}$$

Expand the expression:

$$- {\color{red}{\int{u^{2} \left(1 - u^{2}\right) d u}}} = - {\color{red}{\int{\left(- u^{4} + u^{2}\right)d u}}}$$

Integra término a término:

$$- {\color{red}{\int{\left(- u^{4} + u^{2}\right)d u}}} = - {\color{red}{\left(\int{u^{2} d u} - \int{u^{4} d u}\right)}}$$

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=2$$$:

$$\int{u^{4} d u} - {\color{red}{\int{u^{2} d u}}}=\int{u^{4} d u} - {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=\int{u^{4} d u} - {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

Aplica la regla de la potencia $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=4$$$:

$$- \frac{u^{3}}{3} + {\color{red}{\int{u^{4} d u}}}=- \frac{u^{3}}{3} + {\color{red}{\frac{u^{1 + 4}}{1 + 4}}}=- \frac{u^{3}}{3} + {\color{red}{\left(\frac{u^{5}}{5}\right)}}$$

Recordemos que $$$u=\cos{\left(x \right)}$$$:

$$- \frac{{\color{red}{u}}^{3}}{3} + \frac{{\color{red}{u}}^{5}}{5} = - \frac{{\color{red}{\cos{\left(x \right)}}}^{3}}{3} + \frac{{\color{red}{\cos{\left(x \right)}}}^{5}}{5}$$

Por lo tanto,

$$\int{\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}$$

Añade la constante de integración:

$$\int{\sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)} d x} = \frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}+C$$

$$$\int \sin^{3}{\left(x \right)} \cos^{2}{\left(x \right)}\, dx = \left(\frac{\cos^{5}{\left(x \right)}}{5} - \frac{\cos^{3}{\left(x \right)}}{3}\right) + C$$$A